First slide

Applications of SHM

Question

A simple pendulum has time period T₁. The point of suspension is now moved upward according to equation $y = {kt}^{2}$ where $k = 1 m / \sec^{2}$ . If new time period is T₂ then ratio $\frac{T_{1}^{2}}{T_{2}^{2}}$ will be

Moderate

A

2/3
B

5/6
C

6/5
D

3/2

Solution

$y = {Kt}^{2} \Rightarrow \frac{d^{2} y}{{dt}^{2}} = a_{y} = 2 K = 2 \times 1 = 2 m / s^{2} (∵ K = 1 m / s^{2})$
Now, $T_{1} = 2 π \sqrt{\frac{l}{g}}$ and $T_{2} = 2 π \sqrt{\frac{l}{(g + a_{y})}}$
Dividing, $\frac{T_{1}}{T_{2}} = \sqrt{\frac{g + a_{y}}{g}} \Rightarrow \sqrt{\frac{6}{5}} \Rightarrow \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{6}{5}$

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